C) Calculate the voltage for this cell. V Porous Barrier AI Sn 0.435 M AI3+ 0.00212 M Sn²*

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Please answer C :

## Using the Voltaic Cell Below:

### A) Write the equation for this cell.

### B) Calculate the standard cell potential (E°).

### C) Calculate the voltage for this cell.

### D) Diagram Analysis

In the above diagram:
- **Left Compartment:** Contains an Al (Aluminum) electrode immersed in a solution of \(Al^{3+}\) ions with a concentration of 0.435 M
- **Right Compartment:** Contains a Sn (Tin) electrode immersed in a solution of \(Sn^{2+}\) ions with a concentration of 0.00212 M

Both compartments are separated by a **Porous Barrier**. A voltmeter (V) is connected to measure the potential difference between the two electrodes.

**Anode and Cathode Identification:**
- **Anode:** The electrode where oxidation takes place. Historically, this is the electrode losing electrons.
- **Cathode:** The electrode where reduction takes place. Historically, this is the electrode gaining electrons.

**Flow of Electrons:**
- The electrons flow from the anode to the cathode through the external circuit, which is through the wire connecting the electrodes.

### E) Write the cell diagram. 

---
**Diagram Explanation:**

A schematic representation includes the following components:
- A left half-cell containing the aluminum electrode (labeled "Al") in a solution of \(Al^{3+}\) ions (0.435 M).
- A right half-cell containing the tin electrode (labeled "Sn") in a solution of \(Sn^{2+}\) ions (0.00212 M).
- A porous barrier separating both half-cells to prevent the mixing of different solutions.
- A voltmeter (labeled "V") connecting the aluminum and tin electrodes, measuring the potential difference.
- The flow of electrons is indicated by an arrow pointing from the Aluminum electrode (Anode where oxidation occurs) to the Tin electrode (Cathode where reduction occurs).
Transcribed Image Text:## Using the Voltaic Cell Below: ### A) Write the equation for this cell. ### B) Calculate the standard cell potential (E°). ### C) Calculate the voltage for this cell. ### D) Diagram Analysis In the above diagram: - **Left Compartment:** Contains an Al (Aluminum) electrode immersed in a solution of \(Al^{3+}\) ions with a concentration of 0.435 M - **Right Compartment:** Contains a Sn (Tin) electrode immersed in a solution of \(Sn^{2+}\) ions with a concentration of 0.00212 M Both compartments are separated by a **Porous Barrier**. A voltmeter (V) is connected to measure the potential difference between the two electrodes. **Anode and Cathode Identification:** - **Anode:** The electrode where oxidation takes place. Historically, this is the electrode losing electrons. - **Cathode:** The electrode where reduction takes place. Historically, this is the electrode gaining electrons. **Flow of Electrons:** - The electrons flow from the anode to the cathode through the external circuit, which is through the wire connecting the electrodes. ### E) Write the cell diagram. --- **Diagram Explanation:** A schematic representation includes the following components: - A left half-cell containing the aluminum electrode (labeled "Al") in a solution of \(Al^{3+}\) ions (0.435 M). - A right half-cell containing the tin electrode (labeled "Sn") in a solution of \(Sn^{2+}\) ions (0.00212 M). - A porous barrier separating both half-cells to prevent the mixing of different solutions. - A voltmeter (labeled "V") connecting the aluminum and tin electrodes, measuring the potential difference. - The flow of electrons is indicated by an arrow pointing from the Aluminum electrode (Anode where oxidation occurs) to the Tin electrode (Cathode where reduction occurs).
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